For a compact boson $X\sim X+2\pi R$, a closed spatial circuit of the worldsheet may wind around the target circle:Single-valued target-space wavefunctions quantize the zero-mode momentum as $p=n/R$, $n\in \mathbb{Z}$. In the conventionthe Hamiltonian and worldsheet momentum becomeThe $nw$ term is the zero-mode contribution to closed-string level matching.

Put $\tau ={\tau }_1+i{\tau }_2$ and $q=e^{2\pi i\tau }$. The torus partition function is the traceThe continuum momentum states have density $V/(2\pi )$, so their contribution isEach left-moving oscillator contributes ${\sum }_{N\ge 0}{q}^{nN}=(1-q^n{)}^{-1}$, with the complex conjugate for the right mover. The zero-point factor combines these products into the Dedekind eta functionTherefore the torus partition function of a free boson is

For $n>0$, define occupation numbers $N_n,{\tilde{N}}_n\in {\mathbb{N}}_0$. A basis iswhere $\hat{p}|p;0,0⟩=p|p;0,0⟩$ and positive oscillator modes annihilate the vacuum. Since ${\alpha }_{-n}{\alpha }_n$ has eigenvalue $n{N}_n$, the simultaneous eigenvalues are

The term ${\hat{p}}^{,2}/2$ is the target-space zero-mode kinetic energy. The constant $-1/12$ is the zero-point energy of the left- and right-moving oscillators: each chiral boson contributes $-1/24$. Finally,count the energies of the two independent sets of string oscillators; one excitation of mode $n$ has energy $n$.

Because the Gamma function has simple poles at the nonpositive integers and no zeros, the $s$-dependent numerator $\Gamma (-{\alpha }^{\prime }s/4)$ has poles atorFactorization of a scattering amplitude identifies each pole with an intermediate on-shell state. The Type II superstring mass spectrum therefore contains a massless level at $n=0$ and an infinite equally spaced tower in squared mass for $n\ge 1$. There is no negative-$M^2$ pole, consistently with the absence of a tachyon in Type II superstring theory.

For the Möbius transformation $z^{\prime }=(az+b)/(cz+d)$ with $ad-bc=1$,The power of $|c{z}_i+d|$ contributed by every pair containing $i$ iswhere momentum conservation and the mass-shell condition $p_i^2=4/{\alpha }^{\prime }$ were used. The Koba-Nielsen factor therefore contributes $|c{z}_i+d{|}^4$ at each insertion, exactly cancelling the transformed measure. Hence the remaining integral is $SL(2,\mathbb{C})$ invariant, and division by its volume removes the residual conformal-gauge redundancy.

Split $X^{\mu }=x_0^{\mu }+X^{\prime \mu }$. Integrating the constant mode gives momentum conservation,For the nonzero modes, Wick theorem and the propagator giveThus, up to source-independent numerical normalization, the four-point amplitude isThe product is the Koba-Nielsen factor.

The quadratic worldsheet action makes each $X^{\mu }$ a free boson. Its Green function in the complex plane inverts the operator in the action and uses $\partial \bar{\partial }\log |z-w{|}^2=2\pi {\delta }^{(2)}(z-w)$ with the corresponding convention. The result isAn additive constant is physically irrelevant because it can be absorbed into the zero mode.

The Polyakov action contains the dilaton couplingFor a constant dilaton ${\Phi }_0$, the Gauss-Bonnet theorem gives $S_{\Phi }={\Phi }_0\chi (\Sigma )$, so the path-integral weight contributes $e^{-{\Phi }_0\chi }=g_s^{-\chi }$ with string coupling $g_s=e^{{\Phi }_0}$. A connected closed oriented genus-$g$ worldsheet has $\chi =2-2g$. Including one conventional factor of $g_s$ for each of $n$ external closed-string vertices gives the string genus expansionFor four external states on the sphere, this is $g_s^2$.

The coefficient of the mixed gauge-gravitational anomaly is the sum of all left-handed $U(1)$ charges, with right-handed fields counted as conjugates:For the solution above,More invariantly, $l=-3q$, $u+d=2q$, and $x=-6q$ make the expression vanish identically. Thus the unique cubic-anomaly solution automatically cancels the mixed gauge-gravitational anomaly.

Use the stated rational change of variablesDirect substitution turns the anomaly equation intoWriting $v=a/c$ and $w=b/c$ in lowest common denominator gives the integer equation $a^3+b^3=c^3$. The supplied special case of Fermat's Last Theorem says that one of $a,b$ must vanish. Thus $(v,w)=(1,0)$ or $(0,1)$, which givesThe two signs merely exchange the names of the up- and down-type singlets. Choosing $q=1$ and the conventional sign $y=3$ gives the unique assignment up to overall scaling and that exchange:These are six times the conventional Standard Model hypercharges.

If $q\ne 0$, divide the cubic equation by $q^3$ and define the rational numbersThen every integer anomaly-free assignment supplies a rational solution of

The $[SU(2){]}^{2}U(1)$ condition gives $l=-3q$. Substituting this and $u=q+y$, $d=q-y$ into the cubic anomaly givesUsing $(q+y{)}^3+(q-y{)}^3=2q^3+6q{y}^2$ and multiplying by minus one yields

The $[SU(3){]}^{2}U(1)$ condition gives $u+d=2q$, which is even because all hypercharges are integers. The sum $u+d$ and difference $u-d$ have the same parity, so $u-d$ is even. ThereforeSolving the sum and difference equations gives

Count every right-handed field as a left-handed conjugate, which reverses its $U(1)$ charge and conjugates its non-Abelian representation. The nontrivial local anomaly-cancellation conditions areThe pure $SU(3{)}^3$ anomaly cancels because the two triplets in $Q_L$ balance the conjugates of $u_R$ and $d_R$. There is no perturbative $SU(2{)}^3$ anomaly, and the Witten SU(2) anomaly also vanishes because the three colored quark doublets plus one lepton doublet make four. The mixed gauge-gravitational anomaly is excluded at this stage as requested.

The block-diagonal subgroupis locally $SU(3)\times SU(2)\times U(1)$, with only a finite central quotient distinguishing the global groups. The hypercharge direction in the fundamental representation iswhich is traceless and commutes with the $SU(3)$ and $SU(2)$ blocks.

The canonically normalized $SU(5)$ generator must satisfy $\mathrm{Tr}(T_Y^2)=1/2$. Since $\mathrm{Tr}(Y^2)=5/6$,The unified gauge covariant derivative contains $g_5{T}_Y{B}_{\mu }$, whereas the Standard Model convention contains $g_{Y}Y{B}_{\mu }$. Hence $g_Y=\sqrt{3/5}{g}_5$. The non-Abelian generators already have the canonical normalization, so gauge coupling unification in the SU(5) grand unified theory predicts

Let $K=3(4\pi {)}^2$ and $L=\log (M^2/{\mu }^2)$. Running from $M$ down to $\mu$ givesThe assumed unified boundary condition impliesSubtracting the weak equation from the strong and normalized-hypercharge equations therefore givesEliminating $L$ proves

For $SU(3{)}_c$, one generation contains two color triplets in the quark doublet and the two right-handed color triplets, so $N_f=4N_g$ and there is no colored scalar. Thuswhich permits at most

For $SU(2{)}_L$, each generation contains three colored copies of the quark doublet and one lepton doublet, again giving $N_f=4N_g$. The one complex Higgs doublet gives $N_s=1$, soHence the maximum is

The strong-coupling scale $\Lambda$ is defined as the scale at which the one-loop inverse coupling vanishes. Setting $g^{-2}(\Lambda )=0$ givesEquivalently, in terms of the coupling measured at any scale $\mu$,This dimensional transmutation defines ${\Lambda }_{\mathrm{QCD}}$ for Quantum chromodynamics.

Asymptotic freedom means that the gauge running coupling tends to zero as the renormalization scale tends to infinity. In the stated convention this occurs when $b_0>0$, namelyThe high-energy theory then approaches the Gaussian fixed point and perturbation theory becomes increasingly accurate.